How Can I Order A List Of Connections

I currently have a list of connections stored in a list where each connection is a directed link that connects two points and no point ever links to more than one point or is linke

Solution 1:

Algorithm for a Solution

You're looking for a topological sort algorithm:

from collections import defaultdict

deftopological_sort(dependency_pairs):
    'Sort values subject to dependency constraints'
    num_heads = defaultdict(int)   # num arrows pointing in
    tails = defaultdict(list)      # list of arrows going outfor h, t in dependency_pairs:
        num_heads[t] += 1
        tails[h].append(t)

    ordered = [h for h in tails if h notin num_heads]
    for h in ordered:
        for t in tails[h]:
            num_heads[t] -= 1ifnot num_heads[t]:
                ordered.append(t)
    cyclic = [n for n, heads in num_heads.iteritems() if heads]
    return ordered, cyclic

if __name__ == '__main__':
    connections = [(3, 7), (6, 5), (4, 6), (5, 3), (7, 8), (1, 2), (2, 1)]
    print topological_sort(connections)

Here is the output for your sample data:

([4, 6, 5, 3, 7, 8], [1, 2])

The runtime is linearly proportional to the number of edges (dependency pairs).

HOW IT WORKS

The algorithm is organized around a lookup table called num_heads that keeps a count the number of predecessors (incoming arrows). Consider an example with the following connections: a->h b->g c->f c->h d->i e->d f->b f->g h->d h->e i->b, the counts are:

node  number of incoming edges
----  ------------------------
 a0b2
 c       0
 d       2
 e       1
 f       1  
 g       2
 h       2i1

The algorithm works by "visting" nodes with no predecessors. For example, nodes a and c have no incoming edges, so they are visited first.

Visiting means that the nodes are output and removed from the graph. When a node is visited, we loop over its successors and decrement their incoming count by one.

For example, in visiting node a, we go to its successor h to decrement its incoming count by one (so that h 2 becomes h 1.

Likewise, when visiting node c, we loop over its successors f and h, decrementing their counts by one (so that f 1 becomes f 0 and h 1 becomes h 0).

The nodes f and h no longer have incoming edges, so we repeat the process of outputting them and removing them from the graph until all the nodes have been visited. In the example, the visitation order (the topological sort is):

a c f h e d ib g

If num_heads ever arrives at a state when there are no nodes without incoming edges, then it means there is a cycle that cannot be topologically sorted and the algorithm exits to show the requested results.

Solution 2:

Something like this:

from collections import defaultdict
lis = [ (3, 7), (6, 5), (4, 6), (5, 3), (7, 8), (1, 2), (2, 1) ]
dic = defaultdict(list)

for k,v in lis:
    if v notin dic:
        dic[k].append(v)
    else:
        dic[k].extend([v]+dic[v])
        del dic[v]

for k,v in dic.items():
    for x in v:
        if x in dic and x!=k:
            dic[k].extend(dic[x])
            del dic[x]

print dic
print [[k]+v for k,v in dic.items()]

output:

defaultdict(<type'list'>, {2: [1, 2], 4: [6, 5, 3, 7, 8]})
[[2, 1, 2], [4, 6, 5, 3, 7, 8]]

Solution 3:

I think you can probably do it in O(n) with something like this:

ordered = {}

foreach connection (s,t):
  if t exists in ordered :
     ordered[s] = [t] + ordered[t]
     del ordered[t]
  else:
     ordered[s] = [t]

# Now merge...foreach ordered (s : [t1, t2, ... tN]):
  ordered[s] = [t1, t2, ... tN] + ordered[tN]
  del ordered[tN]

At the end you will be left with something like this, but you can convert that rather easily

ordered= { 
  4 : [6, 5, 3, 7, 8],
  2 : [1, 2]
}